Question

A box contains 30 apples and 40 oranges half of apple and half of orange are rotten if two items are drawn at random what is the probability of either both are rotten or both are apples

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{There}\mathrm{are}30\mathrm{apples}\mathrm{and}40\mathrm{oranges} \\ \mathrm{Rotten}\mathrm{apples}=\frac{1}{2}\times 30=15 \\ \mathrm{Rotten}\mathrm{oranges}=\frac{1}{2}\times 40=20 \\ \mathrm{If}\mathrm{two}\mathrm{items}\mathrm{are}\mathrm{picked} \\ \mathrm{let}\mathrm{A}\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{of}\mathrm{getting}\mathrm{both}\mathrm{items}\mathrm{rotten} \\ \mathrm{P}\left(\mathrm{A}\right)=\frac{{}_2{}^{35}\mathrm{C}}{{}_2{}^{70}\mathrm{C}} \\ =\frac{17}{69} \\ \mathrm{let}\mathrm{B}\mathrm{be}\mathrm{the}\mathrm{event}\mathrm{of}\mathrm{getting}\mathrm{both}\mathrm{items}\mathrm{apples} \\ \mathrm{P}\left(\mathrm{B}\right)=\frac{{}_2{}^{30}\mathrm{C}}{{}_2{}^{70}\mathrm{C}}=\frac{29}{161} \\ \mathrm{P}(\mathrm{A}\cap \mathrm{B})=\frac{15}{70} \\ \mathrm{now},\mathrm{probability}\mathrm{that}\mathrm{either}\mathrm{choosen}\mathrm{items}\mathrm{are}\mathrm{rotten}\mathrm{or}\mathrm{both}\mathrm{are}\mathrm{apples}: \\ \mathrm{P}(\mathrm{A}\cup \mathrm{B})=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}(\mathrm{A}\cap \mathrm{B}) \\ =\frac{17}{69}+\frac{29}{161}-\frac{15}{70} \\ =\frac{205}{966} \end{gathered}$$
Regards